The never-ending finite loop

It's easy to write a loop which looks infinite but in fact completes
quite quickly; for instance, in the C code

for (int i = 1; i > 0; i++);

the variable i starts at 1 and counts upwards "infinitely", but
in fact the loop terminates due to the integer type overflowing and the
value i becoming negative.
A recent
discussion led me to ponder the opposite problem: Can we write a
theoretically finite loop which is nevertheless guaranteed to not
complete?

It turns out that the answer, subject to some qualifications,
is yes. The 48-character line of C99 code

char i,x[99];for(x[98]=i=1;x[98];i++)i*=!++x[i];

takes a finite number of steps to complete; but nevertheless is —
subject to our current understanding of physics and the assumption that
the process responsible for
baryogenesis
can be reversed to cause
proton decay
— guaranteed to never be (non-erronously) completed by a
baryonic computer in the observable universe.

To explain why, let's first examine what this loop does. The statement

i*=!++x[i];

is a wonderful example of C's brevity, but not very informative. Noting
that !x takes the value 0 if x is non-zero and 1 if
x is zero, and that ++ is the pre-increment operator,
we can rewrite this as the considerably more informative

x[i] += 1;
if (x[i] != 0)
i = 0;

Placing this into the context of the original loop, we see that it is
equivalent to the following:

and the inner loop can be easily seen to increment the 98-byte
integer (with characters treated as unsigned base-256
"digits") x[98]x[97]...x[2]x[1]. Exactly how many steps this
takes to complete will depend on the initial contents of the array
x; but it will be between 254 * 256^97 and 256^98, i.e.,
between 1.00 * 10^236 and 1.02 * 10^236.

Having now established that the loop takes a finite number of steps to
complete, we turn to the the second question: Will it ever complete?
The observable universe has an estimated mass of roughly 10^55 kg, or
a mass-energy of roughly 10^72 J; and from Lloyd's famous paper,
Ultimate physical
limits to computation, we know that the number of operations per
second performed by a computer with energy E is at most
4 E / h; so we conclude that a computer in the observable
universe can perform no more than 10^106 operations per second.

How long can such a computer keep computing? Based on our assumption
that the process responsible for baryogenesis can also cause proton
decay, we have from
Adams and Laughlin
an upper bound of roughly 10^41 years on the proton half-life. Even if
our hypothetical universe-computer can continue computing as it
evaporates, the restriction that it is a baryonic computer means
that its mass — and thus its maximum possible processing speed
— will decrease over time, ultimately limiting it to performing
no more than 10^155 operations over its life -- some 81 orders of
magnitude short of of the number of iterations requred to exit our
48-character "finite" loop.

I'll conclude with a request and a challenge. The request: I'm a
computer scientist, not a cosmologist, and it's entirely possible that
there's a glitch in my physics or that I've used the wrong terminology
somewhere. If you know this material better than I do and I've gotten
something wrong, please leave a comment below.

The challenge: I can't see how to write a baryonic-universe-incomputable
finite loop in less than 48 characters of C99 code. Can you do better?